Wind blowing past a flag causes it to “flutter in the breeze.” The frequency of

Question

Wind blowing past a flag causes it to “flutter in the breeze.” The frequency of this fluttering, ?, is assumed to be a function of the wind speed, V, the air density, ?, the acceleration of gravity, g, the length of the flag, l, and the “area density”, ?A (with dimensions of ML-2) of the flag material. It is desired to predict the flutter frequency of a large l = 40 ft flag in a V = 30 ft/s wind. To do this a model flag with l = 4 ft is to be tested in a wind tunnel.
(a) Determine the required area density of the model flag material if the large flag has ?A = 0.006 slugs/ft2.
(b) What wind tunnel velocity is required for testing the model?
(c) If the model flag flutters at 6 Hz, predict the frequency for the large flag.
Answers:
a. 0.0006slugs/ft^2
b. 9.49ft/s
c. 1.90Hz

Explanation / Answer

Flag Length l=40ft
Model flag length lm=4 ft
Velocity, V=30 ft/s
model flag frequency, wm=6 Hz
Area density, pA=0.006 slugs/ft^2
Frequency, w=f(p,g,l,pA)
w=T^-1
v=LT^-1
p=ML^-3
g=LT^-2
l=L
pA=ML^-2
No of ref dimensions=3
No of variables=6
No of pi-terms=6-3=3
wl/g=(v/g,pA/pl)

a)

(pA)m/pmlm=(pA/pl)

(pA)m/lm=pA/l

Model Area density, (pA)m=(pA)/l*(lm)=(0.006/40)*(4/1)=0.0006 slugs/ft^2

(pA)m=0.0006 slugs/ft^2

b)

Vm/(gmlm)=V/(gl)

gm=g

Model velocity, Vm=(lm/l)*(V)=4/40*(30)=9.487 ft/s

Vm=9.487

c)

wm(lm/gm)=wl/g

w=wmlm/l

(6)4/40=1.897 Hz

w=1.9 Hz

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